MATH 222 SECOND SEMESTER CALCULUS
Spring 2011
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Math 222 – 2nd Semester Calculus Lecture notes version 1.7(Spring 2011) This is a self contained set of lecture notes for Math 222. The notes were written by Sigurd Angenent, starting from an extensive collection of notes and problems compiled by Joel Robbin. Some problems were contributed by A.Miller. The LATEX files, as well as the Xfig and Octave files which were used to produce these notes are available at the following web site www.math.wisc.edu/~angenent/Free-Lecture-Notes They are meant to be freely available for non-commercial use, in the sense that “free software” is free. More precisely:
Copyright (c) 2006 Sigurd B. Angenent. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled ”GNU Free Documentation License”.
Contents
Chapter 1: Methods of Integration 3 1. The indefinite integral 3 2. You can always check the answer 4 3. About “+C” 4 4. Standard Integrals 5 5. Method of substitution 5 6. The double angle trick 7 7. Integration by Parts 7 8. Reduction Formulas 9 9. Partial Fraction Expansion 12 10. PROBLEMS 16
Chapter 2: Taylor’s Formulaand Infinite Series 27 11. Taylor Polynomials 27 12. Examples 28 13. Some special Taylor polynomials 32 14. The Remainder Term 32 15. Lagrange’s Formula for the Remainder Term 34 16. The limit as x 0, keeping n fixed 36 → 17. The limit n , keeping x fixed 43 →∞ 18. Convergence of Taylor Series 46 19. Leibniz’ formulas for ln 2 and π/4 48 20. Proof of Lagrange’s formula 49 21. Proof of Theorem 16.8 50 22. PROBLEMS 51
Chapter 3: Complex Numbers and the Complex Exponential 56 23. Complex numbers 56 24. Argument and Absolute Value 57 25. Geometry of Arithmetic 58 26. Applications in Trigonometry 60 27. Calculus of complex valued functions 61 3
28. The Complex Exponential Function 61 29. Complex solutions of polynomial equations 63 30. Other handy things you can do with complex numbers 65 31. PROBLEMS 67
Chapter 4: Differential Equations 72 32. What is a DiffEq? 72 33. First Order Separable Equations 72 34. First Order Linear Equations 73 35. Dynamical Systems and Determinism 75 36. Higher order equations 77 37. Constant Coefficient Linear Homogeneous Equations 78 38. Inhomogeneous Linear Equations 81 39. Variation of Constants 82 40. Applications of Second Order Linear Equations 85 41. PROBLEMS 89
Chapter 5: Vectors 97 42. Introduction to vectors 97 43. Parametric equations for lines and planes 102 44. Vector Bases 104 45. Dot Product 105 46. Cross Product 112 47. A few applications of the cross product 115 48. Notation 118 49. PROBLEMS 118
Chapter 6: Vector Functions and Parametrized Curves 124 50. Parametric Curves 124 51. Examples of parametrized curves 125 52. The derivative of a vector function 127 53. Higher derivatives and product rules 128 54. Interpretation of x′(t) as the velocity vector 129 55. Acceleration and Force 131 56. Tangents and the unit tangent vector 133 57. Sketching a parametric curve 135 58. Length of a curve 137 59. The arclength function 139 60. Graphs in Cartesian and in Polar Coordinates 140 61. PROBLEMS 141
GNU Free Documentation License 148 1. APPLICABILITY AND DEFINITIONS 148 2. VERBATIM COPYING 149 3. COPYING IN QUANTITY 149 4. MODIFICATIONS 149 5. COMBINING DOCUMENTS 150 6. COLLECTIONS OF DOCUMENTS 150 7. AGGREGATION WITH INDEPENDENT WORKS 150 8. TRANSLATION 150 9. TERMINATION 150 10. FUTURE REVISIONS OF THIS LICENSE 151 4
11. RELICENSING 151 5
Chapter 1: Methods of Integration
1. The indefinite integral
We recall some facts about integration from first semester calculus.
1.1. Definition. A function y = F (x) is called an antiderivative of another function y = f(x) if F ′(x)= f(x) for all x.
2 1.2. Example. F1(x)= x is an antiderivative of f(x)=2x. 2 F2(x)= x + 2004 is also an antiderivative of f(x)=2x. 1 G(t)= 2 sin(2t + 1) is an antiderivative of g(t) = cos(2t + 1). The Fundamental Theorem of Calculus states that if a function y = f(x) is continuous on an interval a x b, then there always exists an antiderivative F (x) of f, and one has ≤ ≤ b (1) f(x) dx = F (b) F (a). Za − The best way of computing an integral is often to find an antiderivative F of the given function f, and then to use the Fundamental Theorem (1). How you go about finding an antiderivative F for some given function f is the subject of this chapter. The following notation is commonly used for antiderivates:
(2) F (x)= f(x)dx. Z The integral which appears here does not have the integration bounds a and b. It is called an indefinite integral, as opposed to the integral in (1) which is called a definite integral. It’s important to distinguish between the two kinds of integrals. Here is a list of differences: Indefinite integral Definite integral
b f(x)dx is a function of x. a f(x)dx is a number. R b R By definition f(x)dx is any func- a f(x)dx was defined in terms of tion of x whoseR derivative is f(x). RiemannR sums and can be inter- preted as “area under the graph of y = f(x)”, at least when f(x) > 0. x is not a dummy variable, for exam- x is a dummy variable, for example, ple, 2xdx = x2 + C and 2tdt = 1 1 0 2xdx = 1, and 0 2tdt = 1, so t2 +CRare functions of diffferentR vari- R 1 1 R 0 2xdx = 0 2tdt. ables, so they are not equal. R R 6
2. You can always check the answer
Suppose you want to find an antiderivative of a given function f(x) and after a long and messy computation which you don’t really trust you get an “answer”, F (x). You can then throw away the dubious computation and differentiate the F (x) you had found. If F ′(x) turns out to be equal to f(x), then your F (x) is indeed an antiderivative and your computation isn’t important anymore.
2.1. Example. Suppose we want to find ln x dx. My cousin Bruce says it might be F (x)= x ln x x. Let’s see if he’s right:R − d 1 (x ln x x)= x +1 ln x 1 = ln x. dx − x − Who knows how Bruce thought of this1, but he’s right! We now know that ln xdx = x ln x x + C. − R 3. About “+C”
Let f(x) be a function defined on some interval a x b. If F (x) is an ≤ ≤ antiderivative of f(x) on this interval, then for any constant C the function F˜(x)= F (x)+ C will also be an antiderivative of f(x). So one given function f(x) has many different antiderivatives, obtained by adding different constants to one given antiderivative.
3.1. Theorem. If F1(x) and F2(x) are antiderivatives of the same function f(x) on some interval a x b, then there is a constant C such that F (x) = ≤ ≤ 1 F2(x)+ C. Proof. ′ ′ Consider the difference G(x)= F1(x) F2(x). Then G (x)= F1(x) ′ − − F2(x)= f(x) f(x) = 0, so that G(x) must be constant. Hence F1(x) F2(x)= C for some constant.− −
It follows that there is some ambiguity in the notation f(x) dx. Two functions F1(x) and F2(x) can both equal f(x) dx without equalingR each other. When this happens, they (F1 and F2) differR by a constant. This can sometimes lead to confusing situations, e.g. you can check that
2 sin x cos x dx = sin2 x Z 2 sin x cos x dx = cos2 x Z − are both correct. (Just differentiate the two functions sin2 x and cos2 x!) These − two answers look different until you realize that because of the trig identity sin2 x+ cos2 x = 1 they really only differ by a constant: sin2 x = cos2 x + 1. − To avoid this kind of confusion we will from now on never forget to include the “arbitrary constant +C” in our answer when we compute an antideriv- ative.
1He integrated by parts. 7
4. Standard Integrals
Here is a list of the standard derivatives and hence the standard integrals everyone should know.
f(x) dx = F (x)+ C Z xn+1 xn dx = + C for all n = 1 Z n +1 − 1 dx = ln x + C Z x | | sin x dx = cos x + C Z − cos x dx = sin x + C Z tan x dx = ln cos x + C Z − 1 dx = arctan x + C Z 1+ x2 1 π dx = arcsin x + C (= arccos x + C) Z √1 x2 2 − − dx 1 1 + sin x π π = ln + C for 5. Method of substitution The chain rule says that dF (G(x)) ′ ′ = F (G(x)) G (x), dx so that ′ ′ F (G(x)) G (x) dx = F (G(x)) + C. Z 5.1. Example. Consider the function f(x)= 2x sin(x2 +3). It does not appear in the list of standard integrals we know by heart. But we do notice2 that 2x = d (x2 + 3). So let’s call G(x)= x2 + 3, and F (u)= cos u, then dx − F (G(x)) = cos(x2 + 3) − and dF (G(x)) = sin(x2 + 3) 2x = f(x), dx ′ F ′(G(x)) G (x) | {z } |{z} 2 You will start noticing things like this after doing several examples. 8 so that 2x sin(x2 +3)dx = cos(x2 +3)+ C. Z − The most transparent way of computing an integral by substitution is by in- troducing new variables. Thus to do the integral ′ f(G(x))G (x) dx Z where f(u) = F ′(u), we introduce the substitution u = G(x), and agree to write du = dG(x)= G′(x) dx. Then we get ′ f(G(x))G (x) dx = f(u) du = F (u)+ C. Z Z At the end of the integration we must remember that u really stands for G(x), so that ′ f(G(x))G (x) dx = F (u)+ C = F (G(x)) + C. Z For definite integrals this implies b ′ f(G(x))G (x) dx = F (G(b)) F (G(a)). Za − which you can also write as b G(b) ′ (3) f(G(x))G (x) dx = f(u) du. Za ZG(a) 5.2. Example. [Substitution in a definite integral. ] As an example we com- pute 1 x 2 dx, Z0 1+ x using the substitution u = G(x) =1+ x2. Since du = 2x dx, the associated indefinite integral is 1 1 x dx = 1 du. Z 1+ x2 2 Z u 1 du 1 2 u |{z} | {z } To find the definite integral you must compute the new integration bounds G(0) and G(1) (see equation (3).) If x runs between x = 0 and x = 1, then u = G(x)=1+x2 runs between u =1+02 = 1 and u =1+12 = 2, so the definite integral we must compute is 1 2 x 1 1 2 dx = 2 du, Z0 1+ x Z1 u which is in our list of memorable integrals. So we find 1 2 x 1 2 dx = 1 du = 1 ln u = 1 ln 2. Z 1+ x2 2 Z u 2 1 2 0 1 9 6. The double angle trick If an integral contains sin2 x or cos2 x, then you can remove the squares by using the double angle formulas from trigonometry. Recall that cos2 α sin2 α = cos2α and cos2 α + sin2 α =1, − Adding these two equations gives 1 cos2 α = (cos2α + 1) 2 while substracting them gives 1 sin2 α = (1 cos2α) . 2 − 6.1. Example. The following integral shows up in many contexts, so it is worth knowing: 1 cos2 x dx = (1+cos2x)dx Z 2 Z 1 1 = x + sin 2x + C 2 2 x 1 = + sin 2x + C. 2 4 Since sin 2x = 2 sin x cos x this result can also be written as x 1 cos2 x dx = + sin x cos x + C. Z 2 2 If you don’t want to memorize the double angle formulas, then you can use “Complex Exponentials” to do these and many similar integrals. However, you will have to wait until we are in 28 where this is explained. § 7. Integration by Parts The product rule states d dF (x) dG(x) (F (x)G(x)) = G(x)+ F (x) dx dx dx and therefore, after rearranging terms, dG(x) d dF (x) F (x) = (F (x)G(x)) G(x). dx dx − dx This implies the formula for integration by parts dG(x) dF (x) F (x) dx = F (x)G(x) G(x) dx. Z dx − Z dx 10 7.1. Example – Integrating by parts once. x ex dx = x ex ex 1 dx = xex ex + C. Z − Z − F (x) G′(x) F (x) G(x) G(x) F ′(x) |{z} |{z} |{z} |{z} |{z} |{z} Observe that in this example ex was easy to integrate, while the factor x becomes an easier function when you differentiate it. This is the usual state of affairs when integration by parts works: differentiating one of the factors (F (x)) should simplify the integral, while integrating the other (G′(x)) should not complicate things (too much). Another example: sin x = d ( cos x) so dx − x sin x dx = x( cos x) ( cos x) 1 dx == x cos x + sin x + C. Z − − Z − − 7.2. Example – Repeated Integration by Parts. Sometimes one integra- 2x d 1 2x tion by parts is not enough: since e = dx ( 2 e ) one has e2x e2x x2 e2x dx = x2 2x dx Z 2 − Z 2 F (x) G′(x) |{z} |{z} e2x e2x e2x = x2 2x 2 dx 2 − 4 − Z 4 e2x e2x e2x = x2 2x 2+ C 2 − 4 − 8 1 1 1 = x2e2x xe2x + e2x C 2 − 2 4 − (Be careful with all the minus signs that appear when you integrate by parts.) The same procedure will work whenever you have to integrate P (x)eax dx Z where P (x) is a polynomial, and a is a constant. Each time you integrate by parts, you get this ax ax e e ′ P (x)eax dx = P (x) P (x) dx Z a − Z a 1 1 ′ = P (x)eax P (x)eax dx. a − a Z You have replaced the integral P (x)eax dx with the integral P ′(x)eax dx. This is the same kind of integral, butR it is a little easier since the degreeR of the derivative P ′(x) is less than the degree of P (x). 7.3. Example – My cousin Bruce’s computation. Sometimes the factor G′(x) is “invisible”. Here is how you can get the antiderivative of ln x by integrating 11 by parts: ln x dx = ln x 1 dx Z Z F (x) G′(x) |{z} |{z}1 = ln x x x dx − Z x = x ln x 1 dx − Z = x ln x x + C. − You can do P (x) ln x dx in the same way if P (x) is a polynomial. R 8. Reduction Formulas Consider the integral n ax In = x e dx. Z Integration by parts gives you n 1 ax n−1 1 ax In = x e nx e dx a − Z a 1 n − = xneax xn 1eax dx. a − a Z We haven’t computed the integral, and in fact the integral that we still have to do is of the same kind as the one we started with (integral of xn−1eax instead of xneax). What we have derived is the following reduction formula 1 n ax n In = x e In− (R) a − a 1 which holds for all n. For n = 0 the reduction formula says 1 ax ax 1 ax I0 = e , i.e. e dx = e + C. a Z a When n = 0 the reduction formula tells us that we have to compute In− if we 1 want to find In. The point of a reduction formula is that the same formula also applies to In−1, and In−2, etc., so that after repeated application of the formula we end up with I0, i.e., an integral we know. 8.1. Example. To compute x3eax dx we use the reduction formula three times: R 1 3 I = x3eax I 3 a − a 2 1 3 1 2 = x3eax x2eax I a − a a − a 1 1 3 1 2 1 1 = x3eax x2eax xeax I a − a a − a a − a 0 12 1 ax Insert the known integral I0 = a e + C and simplify the other terms and you get 1 3 6 6 x3eax dx = x3eax x2eax + xeax eax + C. Z a − a2 a3 − a4 8.2. Reduction formula requiring two partial integrations. Consider n Sn = x sin x dx. Z Then for n 2 one has ≥ n n−1 Sn = x cos x + n x cos x dx − Z − − = xn cos x + nxn 1 sin x n(n 1) xn 2 sin x dx. − − − Z Thus we find the reduction formula n n−1 Sn = x cos x + nx sin x n(n 1)Sn− . − − − 2 Each time you use this reduction, the exponent n drops by 2, so in the end you get either S1 or S0, depending on whether you started with an odd or even n. 8.3. A reduction formula where you have to solve for In. We try to compute n In = (sin x) dx Z by a reduction formula. Integrating by parts twice we get n−1 In = (sin x) sin x dx Z − − = (sin x)n 1 cos x ( cos x)(n 1)(sin x)n 2 cos x dx − − Z − − − − = (sin x)n 1 cos x + (n 1) (sin x)n 2 cos2 x dx. − − Z We now use cos2 x =1 sin2 x, which gives − n−1 n−2 n In = (sin x) cos x + (n 1) sin x sin x dx − − Z − n−1 = (sin x) cos x + (n 1)In− (n 1)In. − − 2 − − You can think of this as an equation for In, which, when you solve it tells you n−1 nIn = (sin x) cos x + (n 1)In− − − 2 and thus implies 1 n−1 n 1 In = sin x cos x + − In− . (S) −n n 2 Since we know the integrals 0 I0 = (sin x) dx = dx = x + C and I1 = sin x dx = cos x + C Z Z Z − the reduction formula (S) allows us to calculate In for any n 0. ≥ 13 8.4. A reduction formula which will be handy later. In the next section you will see how the integral of any “rational function” can be transformed into integrals of easier functions, the hardest of which turns out to be dx In = . Z (1 + x2)n When n = 1 this is a standard integral, namely dx I1 = = arctan x + C. Z 1+ x2 When n > 1 integration by parts gives you a reduction formula. Here’s the com- putation: 2 −n In = (1 + x ) dx Z x −n−1 = x ( n) 1+ x2 2x dx (1 + x2)n − Z − x x2 = +2n dx (1 + x2)n Z (1 + x2)n+1 Apply x2 (1 + x2) 1 1 1 = − = (1 + x2)n+1 (1 + x2)n+1 (1 + x2)n − (1 + x2)n+1 to get x2 1 1 dx = dx = In In+1. Z (1 + x2)n+1 Z (1 + x2)n − (1 + x2)n+1 − Our integration by parts therefore told us that x In = +2n In In , (1 + x2)n − +1 which you can solve for In+1. You find the reduction formula 1 x 2n 1 I = + − I . n+1 2n (1 + x2)n 2n n As an example of how you can use it, we start with I1 = arctan x + C, and conclude that dx = I2 = I1+1 Z (1 + x2)2 1 x 2 1 1 = + − I 2 1 (1 + x2)1 2 1 1 x = 1 + 1 arctan x + C. 2 1+ x2 2 14 Apply the reduction formula again, now with n = 2, and you get dx = I3 = I2+1 Z (1 + x2)3 1 x 2 2 1 = + − I 2 2 (1 + x2)2 2 2 2 x x = 1 + 3 1 + 1 arctan x 4 (1 + x2)2 4 2 1+ x2 2 x x = 1 + 3 + 3 arctan x + C. 4 (1 + x2)2 8 1+ x2 8 9. Partial Fraction Expansion A rational function is one which is a ratio of polynomials, n n−1 P (x) pnx + pn−1x + + p1x + p0 f(x)= = d d−1 . Q(x) qdx + qd− x + + q x + q 1 1 0 Such rational functions can always be integrated, and the trick which allows you to do this is called a partial fraction expansion. The whole procedure consists of several steps which are explained in this section. The procedure itself has nothing to do with integration: it’s just a way of rewriting rational functions. It is in fact useful in other situations, such as finding Taylor series (see Part 178 of these notes) and computing “inverse Laplace transforms” (see Math 319.) 9.1. Reduce to a proper rational function. A proper rational function is a rational function P (x)/Q(x) where the degree of P (x) is strictly less than the degree of Q(x). the method of partial fractions only applies to proper rational functions. Fortunately there’s an additional trick for dealing with rational functions that are not proper. If P/Q isn’t proper, i.e. if degree(P ) degree(Q), then you divide P by Q, with result ≥ P (x) R(x) = S(x)+ Q(x) Q(x) where S(x) is the quotient, and R(x) is the remainder after division. In practice you would do a long division to find S(x) and R(x). 9.2. Example. Consider the rational function x3 2x +2 f(x)= − . x2 1 − Here the numerator has degree 3 which is more than the degree of the denominator (which is 2). To apply the method of partial fractions we must first do a division with remainder. One has x +1 = S(x) x2 1 x3 2x +2 − x3 − x −x +2 = R(x) − so that x3 2x +2 x +2 f(x)= − = x + − x2 1 x2 1 − − 15 When we integrate we get x3 2x +2 x +2 − dx = x + − dx Z x2 1 Z x2 1 − − x2 x +2 = + − dx. 2 Z x2 1 − −x+2 The rational function which still have to integrate, namely x2−1 , is proper, i.e. its numerator has lower degree than its denominator. 9.3. Partial Fraction Expansion: The Easy Case. To compute the par- tial fraction expansion of a proper rational function P (x)/Q(x) you must factor the denominator Q(x). Factoring the denominator is a problem as difficult as finding all of its roots; in Math 222 we shall only do problems where the denominator is already factored into linear and quadratic factors, or where this factorization is easy to find. In the easiest partial fractions problems, all the roots of Q(x) are real numbers and distinct, so the denominator is factored into distinct linear factors, say P (x) P (x) = . Q(x) (x a )(x a ) (x an) − 1 − 2 − To integrate this function we find constants A1, A2,...,An so that P (x) A A A = 1 + 2 + + n . (#) Q(x) x a x a x an − 1 − 2 − Then the integral is P (x) dx = A1 ln x a1 + A2 ln x a2 + + An ln x an + C. Z Q(x) | − | | − | | − | One way to find the coefficients Ai in (#) is called the method of equating coefficients. In this method we multiply both sides of (#) with Q(x) = (x − a ) (x an). The result is a polynomial of degree n on both sides. Equating 1 − the coefficients of these polynomial gives a system of n linear equations for A1,..., An. You get the Ai by solving that system of equations. 3 Another much faster way to find the coefficients Ai is the Heaviside trick . 4 Multiply equation (#) by x ai and then plug in x = ai. On the right you are − left with Ai so P (x)(x ai) P (ai) Ai = − = . Q(x) (a a ) (a a − )(a a ) (a a ) x=ai i 1 i i 1 i i+1 i n − − − − 3 Named after Oliver Heaviside, a physicist and electrical engineer in the late 19th and early 20ieth century. 4 More properly, you should take the limit x a . The problem here is that equation (#) → i has x a in the denominator, so that it does not hold for x = a . Therefore you cannot set x − i i equal to a in any equation derived from (#), but you can take the limit x a , which in practice i → i is just as good. 16 x +2 9.4. Previous Example continued. To integrate − we factor the de- x2 1 nominator, − x2 1 = (x 1)(x + 1). − − x +2 The partial fraction expansion of − then is x2 1 − x +2 x +2 A B − = − = + . ( ) x2 1 (x 1)(x + 1) x 1 x +1 † − − − Multiply with (x 1)(x +1) to get − x +2= A(x +1)+ B(x 1)=(A + B)x + (A B). − − − The functions of x on the left and right are equal only if the coefficient of x and the constant term are equal. In other words we must have A + B = 1 and A B =2. − − These are two linear equations for two unknowns A and B, which we now proceed to 1 solve. Adding both equations gives 2A = 1, so that A = 2 ; from the first equation one then finds B = 1 A = 3 . So − − − 2 x +2 1/2 3/2 − = . x2 1 x 1 − x +1 − − Instead, we could also use the Heaviside trick: multiply ( ) with x 1 to get † − x +2 x 1 − = A + B − x +1 x +1 Take the limit x 1 and you find → 1+2 1 − = A, i.e. A = . 1+1 2 Similarly, after multiplying ( ) with x +1 one gets † x +2 x +1 − = A + B, x 1 x 1 − − and letting x 1 you find →− ( 1)+2 3 B = − − = , ( 1) 1 −2 − − as before. Either way, the integral is now easily found, namely, x3 2x +1 x2 x +2 − dx = + x + − dx Z x2 1 2 Z x2 1 − − x2 1/2 3/2 = + x + dx 2 Z x 1 − x +1 − x2 1 3 = + x + ln x 1 ln x +1 + C. 2 2 | − |− 2 | | 17 9.5. Partial Fraction Expansion: The General Case. Buckle up. When the denominator Q(x) contains repeated factors or quadratic factors (or both) the partial fraction decomposition is more complicated. In the most general case the denominator Q(x) can be factored in the form k1 kn 2 ℓ1 2 ℓm (4) Q(x) = (x a ) (x an) (x + b x + c ) (x + bmx + cm) − 1 − 1 1 Here we assume that the factors x a1,..., x an are all different, and we also 2 − 2 − assume that the factors x + b1x + c1,..., x + bmx + cm are all different. It is a theorem from advanced algebra that you can always write the rational function P (x)/Q(x) as a sum of terms like this P (x) A Bx + C (5) = + k + + 2 ℓ + Q(x) (x ai) (x + bjx + cj ) − How did this sum come about? For each linear factor (x a)k in the denominator (4) you get terms − A A A 1 + 2 + + k x a (x a)2 (x a)k − − − in the decomposition. There are as many terms as the exponent of the linear factor that generated them. For each quadratic factor (x2 + bx + c)ℓ you get terms B x + C B x + C B x + C 1 1 + 2 2 + + m m . x2 + bx + c (x2 + bx + c)2 (x2 + bx + c)ℓ Again, there are as many terms as the exponent ℓ with which the quadratic factor appears in the denominator (4). In general, you find the constants A..., B... and C... by the method of equating coefficients. 9.6. Example. To do the integral x2 +3 dx Z x2(x + 1)(x2 + 1)2 apply the method of equating coefficients to the form x2 +3 A A A B x + C B x + C = 1 + 2 + 3 + 1 1 + 2 2 . (EX) x2(x + 1)(x2 + 1)2 x x2 x +1 x2 +1 (x2 + 1)2 Solving this last problem will require solving a system of seven linear equations in the seven unknowns A1, A2, A3,B1, C1,B2, C2. A computer program like Maple can do this easily, but it is a lot of work to do it by hand. In general, the method of equating coefficients requires solving n linear equations in n unknowns where n is the degree of the denominator Q(x). See Problem 104 for a worked example where the coefficients are found. Unfortunately, in the presence of quadratic factors or repeated lin- !! ear factors the Heaviside trick does not give the whole answer; you !! must use the method of equating coefficients. 18 Once you have found the partial fraction decomposition (EX) you still have to integrate the terms which appeared. The first three terms are of the form A(x a)−p dx and they are easy to integrate: − R A dx = A ln x a + C Z x a | − | − and A dx A = + C Z (x a)p (1 p)(x a)p−1 − − − if p> 1. The next, fourth term in (EX) can be written as B1x + C1 x dx dx = B1 dx + C1 Z x2 +1 Z x2 +1 Z x2 +1 B = 1 ln(x2 +1)+ C arctan x + C 2 1 integration const. While these integrals are already not very simple, the integrals Bx + C dx with p> 1 Z (x2 + bx + c)p which can appear are particularly unpleasant. If you really must compute one of these, then complete the square in the denominator so that the integral takes the form Ax + B dx. Z ((x + b)2 + a2)p After the change of variables u = x + b and factoring out constants you have to do the integrals du u du and . Z (u2 + a2)p Z (u2 + a2)p Use the reduction formula we found in example 8.4 to compute this integral. An alternative approach is to use complex numbers (which are on the menu for this semester.) If you allow complex numbers then the quadratic factors x2 + bx + c can be factored, and your partial fraction expansion only contains terms of the form A/(x a)p, although A and a can now be complex numbers. The integrals are then easy,− but the answer has complex numbers in it, and rewriting the answer in terms of real numbers again can be quite involved. 10. PROBLEMS DEFINITE VERSUS INDEFINITE INTEGRALS 1. Compute the following three integrals: 2 2 2 A = x− dx, B = t− dt, C = x− dt. 2. One of the following three integrals is not the same as the other two: 4 4 4 2 2 2 A = x− dx, B = t− dt, C = x− dt. 1 1 1 Which one? BASIC INTEGRALS 19 The following integrals are straightforward provided you know the list of standard antiderivatives. They can be done without using substitution or any other tricks, and you learned them in first semester calculus. π/3 5 4 3. 6x 2x− 7x − − 22. sin tdt x x π/4 +3/x 5 + 4e + 7 dx − π/2 a x 23. (cos θ + 2 sin θ) dθ 4. (x/a + a/x + x + a + ax) dx 0 π/2 3 7 5. √x √x4 + 6ex + 1 dx 24. (cos θ + sin 2θ) dθ − √3 2 − 0 x π tan x 6. 2x + 1 x dx 25. dx 2 2π/3 cos x 0 π/2 cot x 7. (5y4 6y2 + 14) dy 26. dx 3 − π/3 sin x − 3 1 1 √3 6 8. dt 27. dx t2 − t4 1+ x2 1 1 2 t6 t2 0.5 dx 9. − dt 28. 4 2 1 t 0 √1 x − 2 x2 + 1 8 10. dx 29. (1/x) dx √x 1 4 2 ln 6 3 2 11. (x 1) dx 30. 8ex dx − 0 ln 3 2 9 2 12. (x + 1/x) dx 31. 2t dt 1 8 3 e − 3 13. x5 + 2 dx 32. dx 3 e2 x − 1 − 3 14. (x 1)(3x + 2) dx 33. x2 1 dx 1 − 2 | − | − 4 2 15. (√t 2/√t) dt 2 − 34. x x dx 1 1 | − | − 8 1 2 16. √3 r + dr 3 35. (x 2 x ) dx 1 √r 1 − | | − 0 3 2 17. (x + 1) dx 36. (x2 x 1 ) dx 1 − | − | − 0 e x2 + x + 1 2 18. dx x 37. f(x) dx where 1 0 4 9 1 2 x if 0 x< 1, 19. √x + dx f(x)= 5 ≤ √x x , if 1 x 2. 4 ≤ ≤ 1 π 20. √4 x5 + √5 x4 dx 38. f(x) dx where 0 π − 8 x 1 x, if π x 0, 21. − dx f(x)= − ≤ ≤ √3 2 sin x, if 0